Answer is: volume of H₂SO₄ is 42.1 mL.<span>
Chemical reaction: H</span>₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.<span>
c(H</span>₂SO₄) = 0,4567 M = 0,4567 mol/L.<span>
V(NaOH) = 30 mL </span>÷ 1000 mL/L <span>= 0,03 L.
c(NaOH) = 0,321 M = 0,321 mol/L.
n(NaOH) = c(NaOH) · V(NaOH).
n(NaOH) = 0,321 mol/L · 0,030 L.
n(NaOH) = 0,00963 mol.
From chemical reaction: n(H</span>₂SO₄) : n(NaOH) = 1 : 2.<span>
n(H</span>₂SO₄) = 0,01926 mol.<span>
V(H</span>₂SO₄) = n(H₂SO₄) ÷ c(H₂SO₄).<span>
V(H</span>₂SO₄) = 0,01926 mol ÷ 0,4567 mol/L.<span>
V(H</span>₂SO₄<span>) = 0,0421 L = 42,1 mL.</span>
The answer is C. A, B, and C would depend on what type of electron it is.
C - bc as the water boils the cold water particles melt and become the hot water particles!
Answer:
Final T° is 45.5°C
Explanation:
Formula for calorimetry is:
Q = m . C . ΔT , where
ΔT = Final T° - Initial T°
C = Specific heat
m = mass
Let's replace with the data given
305 J = 93.4 g . 0.128 J/g°C . (Final T° - 20°C)
305 J / 93.4 g . 0.128 J/g°C = Final T° - 20°C
25.5°C = Final T° - 20°C → Final T° = 25.5°C + 20°C = 45.5°C
